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I am asked to prove the following theorem:

"Suppose that $G$ is a group with operation $\circ$ and identity $1$. Suppose that $a$ in $G$ and that $x$ is an element such that $a \circ x = a$ or $x \circ a = a$ then $x = 1$."

I am not sure how to address this proof. My thought is to use the definition that there is a unique element in $G$ that is the result of $a \circ b$ for any $a,b$ in $G$. Which means that $a \circ x = a = a \circ 1$ and thus $x=1$. Is this right at all?

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    It is the right plan, but you need to either prove or refer to a proof that you can cancel the $a$ to get from $ax=a1$ to $x=1$. – 2017-02-06
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    @HenningMakholm That's the problem. I am not explicitly allowed to do that. I know that inverses exist defined as: "for every a in G there exists b in G such that ab=1,ba=1" – 2017-02-06
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    x @Kel: Thus, once you have $ax=a1$, multiply from the left by $a^{-1}$ on both sides, and apply associativity. – 2017-02-06

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The property that the product of $a\circ b$ is unique does not lead directly to the result. This property just means that if $a\circ b=x$ and $a\circ b=y$, then $x=y$.

For $a\in G$, there exists an element $b\in G$ such that $$a\circ b=b\circ a=1$$ Suppose that $a\circ x=a$. Then $$a^{-1}\circ (a\circ x)=a^{-1}\circ a$$ By associativity, $$(a^{-1}\circ a)\circ x=a^{-1}\circ a$$ So we get $x=1.$